bug#20954: wc - linux
tele wrote: Maybe we did not understand. I don't want change old definitions but create new option for wc or echo, because this above examples not make logic sense, What would such an option do? ( and it I want fix, however with sed is also fixed ) Your original message asked if echo | wc -l should count 0 lines instead of 1 line. But the echo is going to produce one line and therefore it should be counted. In a later message you wrote using sed to delete blank lines so that only non-blank lines remained to be counted. $ a= ; echo $a | sed '/^\s*$/d' | wc -l 0 $ a=3 ; echo $a | sed '/^\s*$/d' | wc -l 1 Can be added option to wc to fix this problem without use sed in future ? This tells me that you as yet did not understand things yet. :-( I tried to explain this in more detail in my response to that message. The sed command you pulled from stackoverflow.com deletes blank lines. That is a good way to avoid counting blank lines. If I guess at what you are suggesting then it does not make sense to add an option to wc to count only non-blank lines. If you don't want to count blank lines then delete them first. There are an infinite number of possible things to count. There cannot be an infinite number of options implemented. And using sed to delete blank lines is the Right Way To Do Things. however now Iunderstand that they work correctly in accordance with accepted principles. Yes. What is a text line? A text line by definition ends with a newline. This has been standardized to prevent different implementations from implementing it differently and creating portability problems. Therefore all standards compliant implementations must implement it in the same way to prevent portability problems. wc -l in most examples working correct, most? No. wc -l is working correctly in all examples. :-) because it echo give's \n and wc -l count correct. Yes. I mentioned about wc, because for me build option wc -a for echo or echo -m this is not important. Maybe exist hope for example create option m to echo , echo -m which not will from new line, but first line if variable is empty and from new line if is not empty ? example: echo -m | wc -l 0 echo -m e | wc -l 1 The shell is a programming language. If not infinite number then a very large number of possibilities may be implemented by programming them in the shell. All such possibilities should not be coded into specific options. Instead if you have a specific need it should be programmed. Simply write the code that says explicitly what you want to do. There are millions of lines of code written for various tasks. All of those millions of lines should not be turned into specific options. If you want to delete blank lines then simply delete blank lines. This entire discussion feels like an XY problem. Here is a collection of explanations of the XY problem. http://www.perlmonks.org/?node_id=542341 The help-b...@gnu.org mailing list is the right place to follow up but if you wrote there and said what you were trying to do and asking how to do it in the shell people would try to help you there. Bob
bug#20954: wc - linux
tag 20954 + notabug close 20954 thanks Maybe we did not understand. I don't want change old definitions but create new option for wc or echo, because this above examples not make logic sense, ( and it I want fix, however with sed is also fixed ) however now Iunderstand that they work correctly in accordance with accepted principles. # What is a text line? A text line by definition ends with a newline. This has been standardized to prevent # different implementations from implementing it differently and creating portability problems. Therefore # all standards compliant implementations must implement it in the same way to prevent portability # problems. wc -l in most examples working correct, because it echo give's \n and wc -l count correct. I mentioned about wc, because for me build option wc -a for echo or echo -m this is not important. Maybe exist hope for example create option m to echo , echo -m which not will from new line, but first line if variable is empty and from new line if is not empty ? example: echo -m | wc -l 0 echo -m e | wc -l 1
bug#20954: wc - linux
2015-07-01 19:41:00 -0600, Bob Proulx:
[...]
$ a= ; echo $s | wc -l
1
[...]
No. Should be 1. You have forgotten about the newline at the end of
the command. The echo will terminate with a newline.
[...]
Leaving a variable unquoted will also cause the shell to apply
the split+glob operator on it. echo will also do some
transformations on the string (backslash and option processing).
To count the number of bytes in a variable, you can use:
printf %s $var | wc -c
Use ${#var} or
printf %s $var | wc -m
for the number of characters.
GNU wc will not count the bytes that are not part of a valid
character, while GNU bash's ${#var} will count them as one
character:
In a UTF-8 locale:
$ var=$'\x80X\x80\u00e9'
$ printf %s $var | hd
80 58 80 c3 a9|.X...|
0005
$ echo ${#var}
4
$ printf %s $var | wc -c
5
$ printf %s $var | wc -m
2
Above $var contains the 0x80 byte that doesn't form a valid
character, X (0x58), then another 0x80, then é (0xc3 0xa9).
wc -c counts the 5 bytes, wc -m counts X and é, while bash
${#var} counts those plus the 0x80s.
--
Stephane
bug#20954: wc - linux
tag 20954 + notabug close 20954 thanks tele wrote: Hi! Hi! From terminal: $ a= ; echo $s | wc -l 1 Do you mean $a instead of $s? Either way is the same though assuming $s is empty too. - Yes, my mistake :-) Should be 0 , yes ? No. Should be 1. You have forgotten about the newline at the end of the command. The echo will terminate with a newline. You can see this with od. echo | od -tx1 -c 000 0a \n Since this appears to be a usage error I have closed the bug. Please feel free to follow up with more information. We will read it. And we appreciate additional communication! I am simply closing it to keep the accounting straight. Bob # echo gives in new line, echo -n subtracts 1 line, but wc -l can count only from new line, # so if something exist inside first line wc -l can not count. :-( # example: # # $ a=j ; echo $a | wc -l # 1 # # $ a= ; echo $a | wc -l # 1 # # $ a= ; echo -n $a | wc -l # 0 # # $ a=j ; echo -n $a | wc -l # 0 So, $ a= ; echo $a | sed '/^\s*$/d' | wc -l 0 $ a=3 ; echo $a | sed '/^\s*$/d' | wc -l 1 Can be added option to wc to fix this problem without use sed in future ? Thanks for helping :-)
bug#20954: wc - linux
tele wrote: echo gives in new line, Yes. echo -n subtracts 1 line, echo -n is non-portable and shouldn't be used. echo -n suppresses emitting a trailing newline. Note that in both of these cases you are using the shell's internal builtin echo and not the coreutils echo. They behave the same. but wc -l can count only from new line, so if something exist inside first line wc -l can not count. :-( wc -l counts newlines. That is the task that it was constructed to do. That is exactly what it does. No more and no less. What is a text line? A text line by definition ends with a newline. This has been standardized to prevent different implementations from implementing it differently and creating portability problems. Therefore all standards compliant implementations must implement it in the same way to prevent portability problems. example: $ a=j ; echo $a | wc -l 1 I have been wondering. Why are you using a variable here? Using the variable as you are doing is no different than not using the variable. echo j | od -tx1 -c 000 6a 0a j \n There is one newline. That counts as one text line. $ a= ; echo $a | wc -l 1 echo | od -tx1 -c 000 0a \n There is one newline. That counts as one text line. $ a= ; echo -n $a | wc -l 0 echo -n | od -tx1 -c 000 Nothing was emitted. No newlines. Counts as zero lines. But nothing was emitted. Zero characters. od -tx1 -c /dev/null 000 $ a=j ; echo -n $a | wc -l 0 echo -n j | od -tx1 -c 000 6a j That emits one character, the 'j' character. It emits no newlines. Without any newlines at all that is not and cannot be a text line. Without a newline that can only be interpreted as binary data. In any case there were no newlines to count and wc -l counted and reported zero newlines. Instead of echo -n it would be better and portable to use printf instead. printf j | od -tx1 -c 000 6a j Same action in a portable way using printf. Avoid using echo with options. So, $ a= ; echo $a | sed '/^\s*$/d' | wc -l 0 echo | sed '/^\s*$/d' | od -tx1 -c 000 As we previosuly see the echo action will emit one newline character. This is piped to the sed program which will delete that line. Deleting the line is what the sed 'd' action does. Therefore sed does not emit the newline. The text line is deleted. $ a=3 ; echo $a | sed '/^\s*$/d' | wc -l 1 echo 3 | sed '/^\s*$/d' | od -tx1 -c 000 33 0a 3 \n Here the echo emitted two character a '3' and a newline. The sed prgram did not match and therefore did not delete the line. Since it did not delete the line it passed the one text line to wc and wc -l counted the one newline and reported one text line. Can be added option to wc to fix this problem without use sed in future ? Thanks for helping :-) There is no problem to be fixed. And therefore this isn't something that can be fixed in wc. Bob
bug#20954: wc - linux
tag 20954 + notabug close 20954 thanks tele wrote: Hi! Hi! :-) From terminal: $ a= ; echo $s | wc -l 1 Do you mean $a instead of $s? Either way is the same though assuming $s is empty too. Should be 0 , yes ? No. Should be 1. You have forgotten about the newline at the end of the command. The echo will terminate with a newline. You can see this with od. echo | od -tx1 -c 000 0a \n Since this appears to be a usage error I have closed the bug. Please feel free to follow up with more information. We will read it. And we appreciate additional communication! I am simply closing it to keep the accounting straight. :-) Bob
bug#20954: wc - linux
Hi! From terminal: $ a= ; echo $s | wc -l 1 Should be 0 , yes ?
